real-analysis
Prove that there exists $f,g : \mathbb{R}$ to $\mathbb{R}$ such that $f(g(x))$ is strictly increasing and $g(f(x))$ is strictly decreasing.
I tried cases by taking $f(x)$ as an increasing function and $g(x)$ as a decreasing function then I am getting both $f(g(x))$ and $g(f(x))$ as decreasing functions.
Further I took both of them as increasing functions, but none of them are yielding results.
Help
Prove the contrapositive instead: if $f$ is not strictly increasing and not strictly decreasing, then it is not one-to-one.
For example, say there are points $a\lt b\lt c$ such that $f(a)\lt f(b)$ and $f(b)\gt f(c)$. Either $f(a)=f(c)$ (in which case $f$ is not one-to-one), or $f(a)\lt f(c)$, or $f(c)\lt f(a)$.
If $f(a)\lt f(c)\lt f(b)$, then by the Intermediate Value Theorem there exists $d\in (a,b)$ such that $f(d)=f(c)$; hence $f$ is not one-to-one.
Now, there are other possibilities (I made certain assumptions along the way, and you should check what the alternatives are if they are not met).
Aside. Your item 3 sets an unattainable goal. If a continuous function $f$ satisfies $f'=0$ for all but countably many points of $\mathbb R$, then $f$ is a constant function. See Set of zeroes of the derivative of a pathological function.
Here is a more explicit construction than in the post link in comments. Fix $r\in (0,1/2)$. Let $f_0(x)=x$. Inductively define $f_{n+1}$ so that
Here is a piece of this function with $r=1/4$:
and for clarity, here is the family $f_0$,$f_1$,... shown together. All the construction does is replace each line segment of previous step with two; sort of like von Koch snowflake, but less pointy. (It's a monotonic version of the blancmange curve).
To check that the properties hold, you can proceed as follows:
(It may be more enjoyable to prove part 3 using the Law of Large Numbers from probability.)
Best Answer
Here is a construction that uses the axiom of choice. I'm not sure if it can be avoided.
Let $p$ be an increasing bijection ${\mathbb R} \to \mathbb R$ such that $p(0)=0$ and $0$ is the only fixed point of any iterate of $p$ (for example, $p(x)=2017x$ will do).
Let also $q$ be a decreasing bijection ${\mathbb R} \to \mathbb R$ such that $q(0)=0$ and $0$ is the only fixed point of any iterate of $q$ (for example, $q(x)=-2017x$ will do).
The iterates of $p$ form a group under composition which we call $P$. Then $P$ acts on ${\mathbb R}^*$. By the axiom of choice, there is a transversal $T_p\subseteq {\mathbb R}^*$ containing exactly one element from each orbit, then the map ${\mathbb Z} \times T_p \to {\mathbb R}^*, (k,x)\mapsto p^k(x)$ is bijective.
Similarly, there is a $T_q\subseteq {\mathbb R}$ such that ${\mathbb Z} \times T_q \to {\mathbb R}^*, (k,x)\mapsto q^k(x)$ is bijective.
By a well-known result in cardinality theory, since $T_q$ and $\mathbb Z$ are both infinite and $|T_q|>|{\mathbb Z}|$, we have $|{\mathbb Z} \times T_q|=|T_q|$. It follows that there is a bijective map $b:T_q \to T_p$.
Now, define $f$ by $f(0)=0$ and for nonzero $r$, say $r=q^{k}(t)$ with $t\in T_q$, put $f(r)=p^{k}(b(t))$. Similary, define $g$ by $g(0)=0$ and for nonzero $r$, say $r=p^{k}(b(t))$ with $t\in T_q$, put $g(r)=q^{k+1}(t)$.
By construction, one then has $f\circ g=p$ and $g\circ f=q$.